The common definition of the non-negative definite functions is as follows:

Definition 1: A continuous complex-valued function $f(x)$ is called non-negative definite, if for any real numbers $x_1,\dots,x_m$ and complex numbers $\xi_1,\dots,\xi_m$, one has

$$ \sum_{k,j=1}^m f(x_k-x_j) \xi_k \bar{\xi}_j \ge 0\:. $$

See wikipedia for example.

For my use, I need the following more general definition:

Definition 2: A continuous complex-valued function $f(x,y)$ is called non-negative definite, if for any real numbers $x_1,\dots,x_m$ and complex numbers $\xi_1,\dots,\xi_m$, one has

$$ \sum_{k,j=1}^m f(x_k,x_j) \xi_k \bar{\xi}_j \ge 0\:. $$

For the first definition, there is a Characterization by Bochner's theorem, which is discussed in the post.

My question is, for the second definition, is there a characterization like Bochner's theorem? A necessary condition is that $f(x,y)$ is Hermitean in the sense that

$$ f(x,y) = \bar{f}(y,x)\:. $$

Clearly, we need more conditions to guarantee that $f(x,y)$ is non-negative definite. Does any one know a way to determine whether a given function $f(x,y)$ is non-negative definite or not?

Thank you very much for any hints!



Thanks Mikael de la Salle for the useful link. After reading, I still have some problems.

As one might know, Definition 1 above can be extended to distributions. For example, a distribution $F$ is positive-definite, if for every $\psi\in C_c^\infty(R^d)$,

$$ \left(F,\psi *\widetilde{\psi}\right)\ge 0, $$

where $\widetilde{\psi}(x)=\psi(-x)$. So it is natural to think that Definition 2 can also be extended to, say, $C_c^\infty(R^d)\otimes C_c^\infty(R^d)$, or $C_c^\infty(R^{2d})$. The condition in the book Kazhdan's property (T) that the kernel function should be continuous seems too restrictive. Are there some more general statements?

Second EDIT:

Here is a more explicit statement of the problem:

Theorem: Every translation-invariant positive-definite Hermitian bilinear functional $B(\phi,\psi)$ on $C_c^\infty(R^d)$ has the form

$$ B(\phi,\psi)= \int \hat{\phi}(\lambda) \overline{\hat{\psi}(\lambda)} d \mu(\lambda) $$

where $\mu$ is some positive tempered measure and $\hat{\psi}$, $\hat{\phi}$ are the Fourier transforms, respectively, of $\psi$ and $\phi$.

See Gel'fand Volumn 4 P.169 Theorem 6. . My problem is

   What happens if we remove the "translation--invariant" condition?

Thanks a lot!


  • 1
    $\begingroup$ Btw. did you check the classic book on this subject: Harmonic analysis on semigroups by Berg et al.? $\endgroup$ – Suvrit Aug 22 '11 at 17:11
  • $\begingroup$ Not yet. I will have a look. :-) Thanks Suvrit. $\endgroup$ – Anand Aug 22 '11 at 17:14
  • $\begingroup$ @Suvrit, I have a brief look of the book you suggested. It is quite interesting. It deals with the function case (reproducing kernel Hilbert space), which is essentially equivalent to Micael's link below. I am still looking for a more general statement; see the EDIT part of the post. Thank you very much! :-) $\endgroup$ – Anand Aug 22 '11 at 20:04
  • $\begingroup$ I've a hunch that without continuity you will not get an "iff"... $\endgroup$ – Suvrit Aug 22 '11 at 21:40
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    $\begingroup$ @Suvrit Maybe. Btw, the problem of determine whether a continuous function $f(x,y)$ is of positive type is not that easy. Do you know some criteria? if $f(x,y)=f(x-y)$, the we can apply Fourier transform to see whether $\mathcal{F}(f)$ is a positive tempered measure or not. $\endgroup$ – Anand Aug 22 '11 at 21:51

Such a function is called function of positive type. For characterizations, see for example appendix C in the book Kazhdan's property (T) by Bekka, de la Harpe and Valette, available freely on Bachir Bekka's webpage.

  • $\begingroup$ @Mikael de la Salle, Thank you very much. :-) $\endgroup$ – Anand Aug 22 '11 at 14:45
  • $\begingroup$ @Mikael de la Salle, after reading that part, I find the definition there is a little too restrictive. I will add a comment on the EDIT part of the post. $\endgroup$ – Anand Aug 22 '11 at 16:52
  • $\begingroup$ Thanks Mikael de la Salle, I came back to this problem today and suddenly realized that the statement in Appendix C of your reference is so general that it includes my second EDIT. For the reproducing Hilbert space, the original space $X$ can be any topological space, which certainly includes $C_c^\infty(R^d)$. Thank you very much! $\endgroup$ – Anand Jan 26 '12 at 21:43

On a Lie group, every positive-definite distribution is the distributional derivative of some continuous, positive-definite function. See Aarnes, Johan F. Distributions of positive type and representations of Lie groups. Math. Ann. 240 (1979), no. 2, 141–156.

  • $\begingroup$ Thanks Professor Alain Valette, I have had a brief look. The definition of the positive definite in this paper falls in the case Definition 1 above. I am thinking about more general case, the distributional version of Definition 2. :-) $\endgroup$ – Anand Aug 22 '11 at 20:35

In light of Anand's comments on translation invariant kernels of the form $f(x-y)$, I thought I'd mention the class of completely monotone functions $\left((-1)^k f^{(k)}(x) \ge 0\right)$---used in the famous Bernstein and Widder theorem

This theorem allows us to characterize completely monotone functions of the form $k(x,y) = f(x+y)$, by showing that equivalently they must be Laplace transforms of a positive measure.

  • $\begingroup$ Thanks Suvrit. After reading en.wikipedia.org/wiki/… and your suggestion, I find that $k(x,y)=f(x+y)$ should satisfy $x+y\ge 0$, which is quite restricted. Also, we need strong smoothness of the function $f$. Thanks a lot! :-) $\endgroup$ – Anand Aug 23 '11 at 7:12
  • $\begingroup$ Sure; my only point was that this is a well-studied and nicely characterized class of pd functions that are not translation invariant. In general, one can give the useless definition that $k(x,y)=\langle \phi(x), \phi(y)\rangle$ ;-) $\endgroup$ – Suvrit Aug 24 '11 at 17:23

It seems to me that if you drop the translation invariance condition on $B(\phi,\psi)$, then you simply end up with a symmetric quadratic form $B(\phi)=B(\phi,\phi)$. In that case the main simplification I can think of is the spectral representation: $$ B(\phi) = \int_{\operatorname{spec}(B)} ~ \lambda ~ \mathrm{d}\mu_\phi(\lambda) , $$ where $\operatorname{spec}(B)$ is the spectrum of $B$ and $\mathrm{d}\mu_\phi$ is the spectral measure associated to $\phi$.

The spectral theory of quadratic forms is discussed in section VIII.6 of Reed & Simon's Methods of Modern Mathematical Physics (vol. 1). The main theorem is (VIII.15): If $B$ is a semibounded quadratic form, then it is the quadratic form $B(\phi)=(\phi,A\phi)$ of a unique self-adjoint operator $A$. The inner product in your case is the obvious one on $L^2(\mathbb{R})$ and the spectral representation of $B$ is obtained directly from the spectral representation of $A$. Also, since your $B$ is positive semidefinite, the spectrum will also be bounded $\operatorname{spec}(B)\ge 0$.

When $B$ is translation invariant, the unitary transformation that takes $A$ to its diagonal form is automatically the Fourier transform and $\mathrm{d}\mu_\phi(\lambda) = |\hat{\phi}(\lambda)|^2\mathrm{d}\lambda$ (well, up to a measurable reparametrization of $\lambda$ really), where $\hat{\phi}(\lambda)$ is the Fourier transform of $\phi(x)$. In the generic case, short of explicitly diagonalizing $B$ or $A$, I don't see a simple way of obtaining $\mathrm{d}\mu_\phi$.

  • $\begingroup$ Thanks Igor Khavkine. You are working on the measurable function spaces, that's why you need continuous spectrum. If we only consider bilinear functional acting on smooth functions as I mentioned in the second EDIT, a compact operator might be associated and we will probably have point spectra. As a consequence, a similar matrix-like decomposition might be obtained. :-) $\endgroup$ – Anand Aug 23 '11 at 10:50
  • $\begingroup$ By the way, I guess this matrix-like decomposition is the kernel operator of Hilbert-Schwartz type or trace type. :-) $\endgroup$ – Anand Aug 23 '11 at 10:52
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    $\begingroup$ In reply to your first comment, I just want to mention that the theory of spectral decomposition does not presuppose a continuous spectrum. The spectral measure $\mathrm{d}\mu_\phi$ can definitely finite weight on a discrete set. Of course, in your case you may be able to make a priori assumptions about it based on the behavior of the function $f(x,y)$. $\endgroup$ – Igor Khavkine Aug 23 '11 at 13:59
  • $\begingroup$ Yes, I agree. It seems that Section VIII.6 of your reference is what I am looking for. I will read it more carefully. Thanks a lot! :-) $\endgroup$ – Anand Aug 23 '11 at 14:46

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